Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 18G (Galois Theory)

If $a=0$, the splitting field is $K$ and its Galois group is trivial. Otherwise choose a root $\alpha$ and a primitive $n$th root of unity $\zeta \in K$. Every root is ${\zeta }^j\alpha$, so $L=K(\alpha )$. Each automorphism is determined by $\sigma (\alpha )/\alpha \in {\mu }_n$. This is an injective group homomorphism, since the roots of unity are fixed by every automorphism. A subgroup of the cyclic group ${\mu }_n$ is cyclic. Thus $\mathrm{Gal}(L/K)$ is cyclic, with order dividing $n$.

For the indicated cyclic Galois resolvent, reindexing its terms and using $g^m=1$ givesThere exists $w$ with $R(w)\ne 0$ because distinct field automorphisms are linearly independent as functions. A short proof is as follows. Take a nonzero relation with the fewest nonzero terms, normalize one coefficient, evaluate it at $uv$ and subtract a scalar multiple of the relation at $v$. Choose $u$ on which two participating automorphisms differ. This produces a nonzero relation with fewer terms, a contradiction.

Set $\alpha =R(w)\ne 0$. Then $g^j\alpha ={\zeta }^j\alpha$ are $m$ distinct conjugates. Also $g({\alpha }^m)={\alpha }^m$, so $a={\alpha }^m\in K$. The orbit has size $m$, giving $[K(\alpha ):K]=m$ and hence $K(\alpha )=M$. All the roots ${\zeta }^j\alpha$ belong to $M$, soThis also treats $m=1$ with a nonzero $w\in K$.